We split the middle coefficient minus 1 into minus 7 and positive 6 because their sum equals minus 1 and their product equals minus 42, giving the linear binomial factors.
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The polynomial matches the identity template a square – 2ab + b square. Since 16s square is (4s) square and 25t square is (5t) square, it condenses directly into (4s – 5t) square.
This matches the three-term identity where the term containing variable b is negative because both cross-product terms that include b have a minus sign in the expression.
We separate 205 into 200 + 5 and use the addition square identity. Calculating the squares of both numbers and adding their double cross-product gives the final evaluation.
We rewrite the terms as (40 – 6) (40 + 3) and utilize the standard quadratic expansion identity. Computing the numerical components yields the final plain text answer.